A `P`-type semiconductor has acceptor levels `57 meV` above the valence band. The maximum wavelength of light required to create a hole is (Planck's constant `h=6.6xx10^(-34)J-s`)

Options

A. `57 Å`
B. `57xx10^(-3) Å`
C. `217100 Å`
D. `11.61xx10^(-33) Å`

Explanation

To find the maximum wavelength of light required to create a hole in a P-type semiconductor with an acceptor level 57 meV above the valence band, we can follow these steps:

### Step 1: Understand the energy required
The energy required to excite an electron from the valence band to the acceptor level is given as 57 meV (milli-electron volts). We need to convert this energy into joules for our calculations.

### Step 2: Convert meV to joules
1 eV = 1.6 x 10^(-19) J, thus:
\[ 57 \text{ meV} = 57 \times 10^{-3} \text{ eV} = 57 \times 10^{-3} \times 1.6 \times 10^{-19} \text{ J} \]
Calculating this gives:
\[ E = 57 \times 10^{-3} \times 1.6 \times 10^{-19} = 9.12 \times 10^{-21} \text{ J} \]

### Step 3: Use the energy-wavelength relationship
The relationship between energy (E), Planck's constant (h), and wavelength (λ) is given by:
\[ E = \frac{hc}{\lambda} \]
Rearranging this to solve for λ gives:
\[ \lambda = \frac{hc}{E} \]

### Step 4: Substitute the known values
We know:
- Planck's constant \( h = 6.6 \times 10^{-34} \text{ J-s} \)
- Speed of light \( c = 3 \times 10^8 \text{ m/s} \)
- Energy \( E = 9.12 \times 10^{-21} \text{ J} \)

Substituting these values into the equation for λ:
\[ \lambda = \frac{(6.6 \times 10^{-34} \text{ J-s})(3 \times 10^8 \text{ m/s})}{9.12 \times 10^{-21} \text{ J}} \]

### Step 5: Calculate λ
Calculating the numerator:
\[ 6.6 \times 10^{-34} \times 3 \times 10^8 = 1.98 \times 10^{-25} \text{ J-m} \]
Now, dividing by the energy:
\[ \lambda = \frac{1.98 \times 10^{-25}}{9.12 \times 10^{-21}} \approx 2.17 \times 10^{-7} \text{ m} \]
Converting meters to angstroms (1 m = \( 10^{10} \) angstroms):
\[ \lambda \approx 2.17 \times 10^{-7} \text{ m} \times 10^{10} \text{ A/m} \approx 2170 \text{ Å} \]

### Final Answer
The maximum wavelength of light required to create a hole in the P-type semiconductor is approximately **2170 Å**.

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